Let `f(x)=lim_( n to oo)(cosx)/(1+(tan^(-1)x)^(n))`. Then the value of `int_(o)^(oo)f(x)dx` is equal to

Options

A. `cos (tan 1)`
B. `sin(tan1)`
C. `tan(tan1)`
D. none of these

Explanation

To solve the problem, we need to evaluate the function \( f(x) \) and then compute the integral \( \int_0^\infty f(x) \, dx \).

### Step 1: Evaluate \( f(x) \)
We start with the definition of \( f(x) \):
\[ f(x) = \lim_{n \to \infty} \frac{\cos x}{1 + (\tan^{-1} x)^n} \]

### Step 2: Analyze the behavior of \( \tan^{-1} x \)
- For \( x > 1 \), \( \tan^{-1} x \) approaches \( \frac{\pi}{4} \) (which is a finite value).
- For \( x < 1 \), \( \tan^{-1} x \) approaches \( 0 \).

### Step 3: Determine the limit
- If \( x > 1 \), then \( (\tan^{-1} x)^n \to \infty \) as \( n \to \infty \). Thus, \( f(x) \) becomes:
\[ f(x) = \lim_{n \to \infty} \frac{\cos x}{1 + \infty} = 0 \]
- If \( 0 < x < 1 \), then \( \tan^{-1} x \) is a finite value less than \( 1 \), and \( (\tan^{-1} x)^n \to 0 \) as \( n \to \infty \). Thus, \( f(x) \) becomes:
\[ f(x) = \frac{\cos x}{1 + 0} = \cos x \]
- If \( x = 0 \), \( f(0) = \cos(0) = 1 \).

### Step 4: Combine the results
We can summarize:
\[ f(x) = \begin{cases} \cos x & \text{if } 0 < x < 1 \\ 0 & \text{if } x \geq 1 \end{cases} \]

### Step 5: Set up the integral
Now we compute the integral:
\[ \int_0^\infty f(x) \, dx = \int_0^1 \cos x \, dx + \int_1^\infty 0 \, dx \]
The second integral is zero, so we only need to evaluate:
\[ \int_0^1 \cos x \, dx \]

### Step 6: Evaluate the integral
The integral of \( \cos x \) is:
\[ \int \cos x \, dx = \sin x \]
Thus,
\[ \int_0^1 \cos x \, dx = \sin(1) - \sin(0) = \sin(1) - 0 = \sin(1) \]

### Final Answer
Therefore, the value of \( \int_0^\infty f(x) \, dx \) is:
\[ \sin(1) \]